Problem: For a polynomial $p(x),$ define its munificence as the maximum value of $|p(x)|$ on the interval $-1 \le x \le 1.$  For example, the munificence of the polynomial $p(x) = -x^2 + 3x - 17$ is 21, since the maximum value of $|-x^2 + 3x - 17|$ for $-1 \le x \le 1$ is 21, occurring at $x = -1.$

Find the smallest possible munificence of a monic quadratic polynomial.
Answer: Let $f(x) = x^2 + bx + c,$ and let $M$ be the munificence of $f(x).$  Then $|f(-1)| \le M,$ $|f(0)| \le M$ and $|f(1)| \le M.$  These lead to
\begin{align*}
|1 - b + c| &\le M, \\
|c| &\le M, \\
|1 + b + c| & \le M.
\end{align*}Then by Triangle Inequality,
\begin{align*}
4M &= |1 - b + c| + 2|c| + |1 + b + c| \\
&= |1 - b + c| + 2|-c| + |1 + b + c| \\
&\ge |(1 - b + c) + 2(-c) + (1 + b + c)| \\
&= 2.
\end{align*}Hence, $M \ge \frac{1}{2}.$

Consider the quadratic $f(x) = x^2 - \frac{1}{2}.$  Then
\[-\frac{1}{2} \le x^2 - \frac{1}{2} \le \frac{1}{2}\]for $-1 \le x \le 1,$ and $|f(-1)| = |f(0)| = |f(1)| = \frac{1}{2},$ so munificence of $f(x)$ is $\frac{1}{2}.$

Therefore, the smallest possible munificence of a monic quadratic polynomial is $\boxed{\frac{1}{2}}.$